1. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
2-1
Lecture slides to accompany
Engineering Economy
7th
edition
Leland Blank
Anthony Tarquin
Chapter 2Chapter 2
Factors: How TimeFactors: How Time
and Interest Affectand Interest Affect
MoneyMoney

2. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
2-2
LEARNING OUTCOMESLEARNING OUTCOMES
1. F/P and P/F Factors
2. P/A and A/P Factors
3. F/A and A/F Factors
4. Factor Values
5. Arithmetic Gradient
6. Geometric Gradient
7. Find i or n

3. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
2-3
Single Payment Factors (F/P and P/F)Single Payment Factors (F/P and P/F)
Single payment factors involve only P and F. Cash flow diagrams are as follows:
F = P(1 + i ) n P = F[1 / (1 + i ) n
]
Formulas are as follows:
Terms in parentheses or brackets are called factors. Values are in tables for i and n values
Factors are represented in standard factor notation such as (F/P,i,n),
where letter to left of slash is what is sought; letter to right represents what is given

4. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
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F/P and P/F for SpreadsheetsF/P and P/F for Spreadsheets
Future value F is calculated using FV function:
= FV(i%,n,,P)
Present value P is calculated using PV function:
= PV(i%,n,,F)
Note the use of double commas in each function

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Example: Finding Future ValueExample: Finding Future Value
A person deposits $5000 into an account which pays interest at a rate of 8%
per year. The amount in the account after 10 years is closest to:
(A) $2,792 (B) $9,000 (C) $10,795 (D) $12,165
The cash flow diagram is:
Solution:
F = P(F/P,i,n )
= 5000(F/P,8%,10 )
= $10,794.50
Answer is (C)
= 5000(2.1589)

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Example: Finding Present ValueExample: Finding Present Value
A small company wants to make a single deposit now so it will have enough money to
purchase a backhoe costing $50,000 five years from now. If the account will earn
interest of 10% per year, the amount that must be deposited now is nearest to:
(A) $10,000 (B) $ 31,050 (C) $ 33,250 (D) $319,160
The cash flow diagram is: Solution:
P = F(P/F,i,n )
= 50,000(P/F,10%,5 )
= 50,000(0.6209)
= $31,045
Answer is (B)

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Uniform Series Involving P/A and A/PUniform Series Involving P/A and A/P
The cash flow diagrams are:
Standard Factor NotationP = A(P/A,i,n) A = P(A/P,i,n)
Note: P is one period Ahead of first A value
(1) Cash flow occurs in consecutive interest periods
The uniform series factors that involve P and A are derived as follows:
(2) Cash flow amount is same in each interest period

8. Example: Uniform Series Involving P/AExample: Uniform Series Involving P/A
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2-8
A chemical engineer believes that by modifying the structure of a certain water
treatment polymer, his company would earn an extra $5000 per year. At an interest
rate of 10% per year, how much could the company afford to spend now to just
break even over a 5 year project period?
(A) $11,170 (B) 13,640 (C) $15,300 (D) $18,950
The cash flow diagram is as follows:
P = 5000(P/A,10%,5)
= 5000(3.7908)
= $18,954
Answer is (D)
Solution:

9. Uniform Series Involving F/A and A/FUniform Series Involving F/A and A/F
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(1) Cash flow occurs in consecutive interest periods
The uniform series factors that involve F and A are derived as follows:
(2) Last cash flow occurs in same period as F
Note: F takes place in the same period as last A
Cash flow diagrams are:
Standard Factor NotationF = A(F/A,i,n) A = F(A/F,i,n)

10. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
2-10
Example: Uniform Series Involving F/AExample: Uniform Series Involving F/A
An industrial engineer made a modification to a chip manufacturing
process that will save her company $10,000 per year. At an interest
rate of 8% per year, how much will the savings amount to in 7 years?
(A) $45,300 (B) $68,500 (C) $89,228 (D) $151,500
The cash flow diagram is:
A = $10,000
F = ?
i = 8%
0 1 2 3 4 5 6 7
Solution:
F = 10,000(F/A,8%,7)
= 10,000(8.9228)
= $89,228
Answer is (C)

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Factor Values for Untabulated i or nFactor Values for Untabulated i or n
3 ways to find factor values for untabulated i or n values
Use formula
Use spreadsheet function with corresponding P, F, or A value set to 1
Linearly interpolate in interest tables
Formula or spreadsheet function is fast and accurate
Interpolation is only approximate

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2-12
Example: Untabulated iExample: Untabulated i
Determine the value for (F/P, 8.3%,10)
Formula: F = (1 + 0.083)10
= 2.2197
Spreadsheet: = FV(8.3%,10,,1) = 2.2197
Interpolation: 8% ------ 2.1589
8.3% ------ x
9% ------ 2.3674
x = 2.1589 + [(8.3 - 8.0)/(9.0 - 8.0)][2.3674 – 2.1589]
= 2.2215
Absolute Error = 2.2215 – 2.2197 = 0.0018
OK
OK
(Too high)

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2-13
Arithmetic GradientsArithmetic Gradients
Arithmetic gradients change by the same amount each period
The cash flow diagram for the PG
of an arithmetic gradient is: G starts betweenG starts between periods 1 and 2periods 1 and 2
(not between 0 and 1)(not between 0 and 1)
This is because cash flow in year 1 is
usually not equal to G and is handled
separately as a base amount
(shown on next slide)
Note that PNote that PGG isis locatedlocated TwoTwo PeriodsPeriods
AheadAhead of the first change that is equalof the first change that is equal
to Gto G
Standard factor notation is
PG = G(P/G,i,n)

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Typical Arithmetic Gradient Cash FlowTypical Arithmetic Gradient Cash Flow
PT = ?
i = 10%
0 1 2 3 4 5
400
450
500
550
600
PA = ?
i = 10%
0 1 2 3 4 5
400 400 400 400 400
PG = ?
i = 10%
0 1 2 3 4 5
50
100
150
200
+
This diagram = this base amount plus this gradient
PA = 400(P/A,10%,5) PG = 50(P/G,10%,5)
PT = PA + PG = 400(P/A,10%,5) + 50(P/G,10%,5)
Amount
in year 1
is base
amount
Amount in year 1
is base amount

15. Converting Arithmetic Gradient to AConverting Arithmetic Gradient to A
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2-15
i = 10%
0 1 2 3 4 5
G
2G
3G
4G
i = 10%
0 1 2 3 4 5
A = ?
Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n)
General equation when base amount is involved is
A = base amount + G(A/G,i,n)
0 1 2 3 4 5
G
2G
3G
4G
For decreasing gradients,
change plus sign to minus
A = base amount - G(A/G,i,n)

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2-16
Example: Arithmetic GradientExample: Arithmetic Gradient
= 400(3.6048) + 30(6.3970)
= $1,633.83
Answer is (B)
PT = 400(P/A,12%,5) + 30(P/G,12%,5)
The cash flow could also be converted
into an A value as follows:
A = 400 + 30(A/G,12%,5)
= 400 + 30(1.7746)
= $453.24
Solution:

17. Geometric GradientsGeometric Gradients
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2-17
Geometric gradients change by the same percentage each period
0
1 2 3 n
A1
A 1(1+g)1
4
A 1(1+g)2
A 1(1+g)n-1
Pg = ?
There are no tables for geometric factors
Use following equation for g ≠ i:
Pg = A1{1- [(1+g)/(1+i)]n
}/(i-g)
where: A1 = cash flow in period 1
g = rate of increase
If g = i, Pg = A1n/(1+i)
Note: If g is negative, change signs in front of both g values
Cash flow diagram for present worth
of geometric gradient
Note: g starts between
periods 1 and 2

18. Example: Geometric GradientExample: Geometric Gradient
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2-18
0
1 2 3 10
1000
1070
4
1145
1838
Pg = ? Solution:
Pg = 1000[1-(1+0.07/1+0.12)10
]/(0.12-0.07)
= $7,333
Answer is (b)
g = 7%
i = 12%
To find A, multiply Pg by (A/P,12%,10)

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2-19
Unknown Interest Rate i
(Usually requires a trial and error solution or interpolation in interest tables)
Can use either the P/A or A/P factor. Using A/P:Solution:
60,000(A/P,i%,10) = 16,000
(A/P,i%,10) = 0.26667
From A/P column at n = 10 in the interest tables, i is between 22% and 24% Answer is (d)
Procedure: Set up equation with all symbols involved and solve for i

20. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
2-20
Unknown Recovery Period nUnknown Recovery Period n
Unknown recovery period problems involve solving for n,
given i and 2 other values (P, F, or A)
(Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables)
Procedure: Set up equation with all symbols involved and solve for n
A contractor purchased equipment for $60,000 that provided income of $8,000
per year. At an interest rate of 10% per year, the length of time required to recover
the investment was closest to:
(a) 10 years (b) 12 years (c) 15 years (d) 18 years
Can use either the P/A or A/P factor. Using A/P:Solution:
60,000(A/P,10%,n) = 8,000
(A/P,10%,n) = 0.13333
From A/P column in i = 10% interest tables, n is between 14 and 15 years Answer is (c)

21. Summary of Important PointsSummary of Important Points
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2-21
In P/A and A/P factors, P is one period ahead of first A
In F/A and A/F factors, F is in same period as last A
To find untabulated factor values, best way is to use formula or spreadsheet
For arithmetic gradients, gradient G starts between periods 1 and 2
Arithmetic gradients have 2 parts, base amount (year 1) and gradient amount
For geometric gradients, gradient g starts been periods 1 and 2
In geometric gradient formula, A1 is amount in period 1
To find unknown i or n, set up equation involving all terms and solve for i or n